Integral Calculus Notes

Indefinite Integrals and U-Substitution

Basic Integral Formulas

  • ∫sec⁡(x)tan⁡(x)dx=sec⁡(x)+C\int \sec(x) \tan(x) dx = \sec(x) + C
  • ∫csc⁡(x)cot⁡(x)dx=−csc⁡(x)+C\int \csc(x) \cot(x) dx = -\csc(x) + C

Inverse Trigonometric Integrals

  • ∫dx1−x2=sin⁡−1(x)+C\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}(x) + C
  • ∫−dx1−x2=cos⁡−1(x)+C\int \frac{-dx}{\sqrt{1 - x^2}} = \cos^{-1}(x) + C

Power Rule Integrals

  • The general power rule for integration:
    ∫[g(x)]ng′(x)dx=[g(x)]n+1n+1+C\int [g(x)]^n g'(x) dx = \frac{[g(x)]^{n+1}}{n+1} + C where n≠−1n \neq -1
  • If g(x)=xg(x) = x, this simplifies to the basic power rule:
    ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C

U-Substitution

  • U-substitution is useful when the integrand contains a composite function f(g(x))f(g(x)) and its derivative g′(x)g'(x). The goal is to transform the integral into a simpler form by substituting u=g(x)u = g(x) and du=g′(x)dxdu = g'(x) dx.

Examples

Example 1: Power Rule with Trigonometric Function
  • Evaluate ∫sin⁡2(x)cos⁡(x)dx\int \sin^2(x) \cos(x) dx
  • Here, g(x)=sin⁡(x)g(x) = \sin(x), and g′(x)=cos⁡(x)g'(x) = \cos(x).
  • Applying the power rule:
    ∫sin⁡2(x)cos⁡(x)dx=sin⁡3(x)3+C\int \sin^2(x) \cos(x) dx = \frac{\sin^3(x)}{3} + C
Example 2: Power Rule with Polynomial Function
  • Evaluate ∫(2x2+3x)2(4x+3)dx\int (2x^2 + 3x)^2 (4x + 3) dx
  • Here, g(x)=2x2+3xg(x) = 2x^2 + 3x, and g′(x)=4x+3g'(x) = 4x + 3.
  • Applying the power rule:
    ∫(2x2+3x)2(4x+3)dx=(2x2+3x)33+C\int (2x^2 + 3x)^2 (4x + 3) dx = \frac{(2x^2 + 3x)^3}{3} + C
Example 3: U-Substitution with Exponential Function
  • Evaluate ∫ex2xdx\int e^{x^2} x dx
  • Let u=x2u = x^2, then du=2xdxdu = 2x dx, and xdx=12dux dx = \frac{1}{2} du
  • The integral becomes:
    ∫ex2xdx=12∫eudu=12eu+C=12ex2+C\int e^{x^2} x dx = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C
Explanation of Exponential Integral
  • The integral ∫eg(x)g′(x)dx=eg(x)+C\int e^{g(x)} g'(x) dx = e^{g(x)} + C
  • Consider the derivative of F(x)=ex2+CF(x) = e^{x^2} + C:
    dFdx=ddx(ex2+C)=ex2(2x)\frac{dF}{dx} = \frac{d}{dx}(e^{x^2} + C) = e^{x^2} (2x)
  • This verifies that ex2e^{x^2} is the antiderivative of ex2(2x)e^{x^2} (2x).
Example 4: U-Substitution with Inverse Sine
  • Evaluate ∫2x1−(x2)2dx\int \frac{2x}{\sqrt{1 - (x^2)^2}} dx
  • Let g(x)=x2g(x) = x^2, so the integral resembles the form ∫g′(x)1−[g(x)]2dx\int \frac{g'(x)}{\sqrt{1 - [g(x)]^2}} dx
  • Therefore, ∫2x1−(x2)2dx=sin⁡−1(x2)+C\int \frac{2x}{\sqrt{1 - (x^2)^2}} dx = \sin^{-1}(x^2) + C
Example 5: U-Substitution with Inverse Tangent
  • Evaluate ∫2x1+x4dx\int \frac{2x}{1 + x^4} dx
  • Rewrite the integral as ∫2x1+(x2)2dx\int \frac{2x}{1 + (x^2)^2} dx
  • Let g(x)=x2g(x) = x^2, then g′(x)=2xg'(x) = 2x
  • Using the formula ∫g′(x)1+[g(x)]2dx=tan⁡−1(g(x))+C\int \frac{g'(x)}{1 + [g(x)]^2} dx = \tan^{-1}(g(x)) + C:
    ∫2x1+x4dx=tan⁡−1(x2)+C\int \frac{2x}{1 + x^4} dx = \tan^{-1}(x^2) + C

General Cases

  • Type 1: Integrals where g(x)=xg(x) = x is a specific (simpler) case.

  • Type 2: Integrals involving a more general function g(x)g(x).

  • ∫dxxx2−1=sec⁡−1(∣x∣)+C\int \frac{dx}{x \sqrt{x^2 - 1}} = \sec^{-1}(|x|) + C

  • For a general case, it becomes ∫g′(x)g(x)[g(x)]2−1dx=sec⁡−1(∣g(x)∣)+C\int \frac{g'(x)}{g(x) \sqrt{[g(x)]^2 - 1}} dx = \sec^{-1}(|g(x)|) + C

Example 6: Inverse Tangent with Trigonometric Function
  • Evaluate ∫cos⁡(x)1+sin⁡2(x)dx\int \frac{\cos(x)}{1 + \sin^2(x)} dx
  • Let g(x)=sin⁡(x)g(x) = \sin(x), so g′(x)=cos⁡(x)g'(x) = \cos(x)
  • Using the formula ∫g′(x)1+[g(x)]2dx=tan⁡−1(g(x))+C\int \frac{g'(x)}{1 + [g(x)]^2} dx = \tan^{-1}(g(x)) + C:
    ∫cos⁡(x)1+sin⁡2(x)dx=tan⁡−1(sin⁡(x))+C\int \frac{\cos(x)}{1 + \sin^2(x)} dx = \tan^{-1}(\sin(x)) + C
Example 7: Inverse Sine with Exponential Function
  • Evaluate ∫7xln⁡(7)1−(7x)2dx\int \frac{7^x \ln(7)}{\sqrt{1 - (7^x)^2}} dx
  • Let g(x)=7xg(x) = 7^x, so g′(x)=7xln⁡(7)g'(x) = 7^x \ln(7)
  • Using the formula ∫g′(x)1−[g(x)]2dx=sin⁡−1(g(x))+C\int \frac{g'(x)}{\sqrt{1 - [g(x)]^2}} dx = \sin^{-1}(g(x)) + C:
    ∫7xln⁡(7)1−(7x)2dx=sin⁡−1(7x)+C\int \frac{7^x \ln(7)}{\sqrt{1 - (7^x)^2}} dx = \sin^{-1}(7^x) + C
Example 8: Logarithmic Integral
  • Evaluate ∫11+xdx\int \frac{1}{1 + x} dx
  • Recognize this as ∫g′(x)g(x)dx=ln⁡∣g(x)∣+C\int \frac{g'(x)}{g(x)} dx = \ln|g(x)| + C:
    ∫11+xdx=ln⁡∣1+x∣+C\int \frac{1}{1 + x} dx = \ln|1 + x| + C